Bernstein's Theorem on Unique Global Solution to y''=F(x,y,y')/Lemma 3
Lemma for Bernstein's Theorem on Unique Global Solution to $y'' = \map F {x, y, y'}$
Let $F$ and its partial derivatives $F_y, F_{y'}$ be real functions, defined on the closed interval $I = \closedint a b$.
Let $F, F_y, F_{y'} $ be continuous at every point $\tuple {x, y}$ for all finite $y'$.
Suppose there exists a constant $k > 0$ such that:
- $\map {F_y} {x, y, y'} > k$
Suppose there exist real functions $\alpha = \map \alpha {x, y} \ge 0$, $\beta = \map \beta {x, y}\ge 0$ bounded in every bounded region of the plane such that:
- $\size {\map F {x, y, y'} } \le \alpha y'^2 + \beta$
Suppose that for $x \in I$:
- $\map {y''} x = \map F {x, y, y'}$
where:
- $\map y a = a_1$
- $\map y c = c_1$
Then:
- $\forall x \in I: \exists M \in \R: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le M$
Proof
Let $\AA$ and $\BB$ be the least upper bounds of $\map \alpha {x, y}$ and $\map \beta {x, y}$ respectively in the rectangle $a \le x \le c$, $\size y \le m + max \set {\size a_1, \size c_1}$
where:
- $m = \dfrac 1 k \max \limits_{a \mathop \le x \mathop \le b} \size {\map F {x, \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a}, \dfrac {c_1 - a_1} {c - a} } }$
Suppose that $\AA \ge 1$.
Let $u, v$ be real functions such that:
\(\text {(3)}: \quad\) | \(\ds \map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + m\) | \(=\) | \(\ds \dfrac {\Ln u} {2 \AA}\) | |||||||||||
\(\text {(4)}: \quad\) | \(\ds -\map y x + \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} + m\) | \(=\) | \(\ds \dfrac {\Ln v} {2 \AA}\) |
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From Lemma 2, for $x \in I$ the left hand sides of $(3)$ and $(4)$ are not negative.
Thus:
- $\forall x \in I: u, v \ge 1$
Differentiate equations $(3)$ and $(4)$ with respect to $x$:
\(\text {(5)}: \quad\) | \(\ds \map {y'} x - \dfrac {c_1 - a_1} {c - a}\) | \(=\) | \(\ds \dfrac {u'} {2 \AA u}\) | |||||||||||
\(\text {(6)}: \quad\) | \(\ds -\map {y'} x + \dfrac {c_1 - a_1} {c - a}\) | \(=\) | \(\ds \dfrac {v'} {2 \AA v}\) |
Differentiate again:
\(\text {(7)}: \quad\) | \(\ds \map {y''} x\) | \(=\) | \(\ds \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2}\) | |||||||||||
\(\text {(8)}: \quad\) | \(\ds -\map {y''} x\) | \(=\) | \(\ds \dfrac {v''} {2 \AA v} - \dfrac {v'^2} {2 \AA v^2}\) |
By assumption:
\(\ds \size {\map F {x, y, y'} }\) | \(=\) | \(\ds \size {\map {y''} x}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map \alpha {x, y} \map {y'^2} x + \map \beta {x, y}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \AA \map {y'^2} x + \BB\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \AA \map {y'^2} x + \BB\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \AA {\paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2} + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2 - 2 \map {y'} x \dfrac {c_1 - a_1} {c - a} + \BB\) | ||||||||||||
\(\text {(9)}: \quad\) | \(\ds \) | \(\le\) | \(\ds 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1\) |
where:
- $\BB_1 = \BB + 2 \AA \paren {\dfrac {c_1 - a_1} {c - a} }^2$
Then:
\(\ds y''\) | \(=\) | \(\ds \dfrac {u''} {2 \AA u} - \dfrac {u'^2} {2 \AA u^2}\) | Equation $(7)$ | |||||||||||
\(\ds \) | \(\ge\) | \(\ds -2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 - \BB_1\) | Inequality $(9)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -2 \AA \paren {\dfrac {u'} {2 \AA u} }^2 - \BB_1\) | Equation $(5)$ |
Multiply the inequality by $2 \AA u$ and simplify:
\(\ds u''\) | \(\ge\) | \(\ds -2 \AA \BB_1 u\) |
Similarly:
\(\ds y''\) | \(=\) | \(\ds - \dfrac {v''} {2 \AA v} + \dfrac {v'^2} {2 \AA v^2}\) | Equation $(8)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \AA \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2 + \BB_1\) | Inequality $(9)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 \AA \paren {\dfrac {v'} {2 \AA v} }^2 + \BB_1\) | Equation $(6)$ |
Multiply the inequality by $-2 \AA v$ and simplify:
\(\ds v''\) | \(\ge\) | \(\ds -2 \AA \BB_1 v\) |
Note that $\map u a = \map u c$ and $\map v a = \map v c$.
From Intermediate Value Theorem it follows that
- $\exists K \subset I: \forall x_0 \in K: \map {y'} {x_0} - \dfrac {c_1 - a_1} {c - a} = 0$
Points $x_0$ divide $I$ into subintervals.
Due to $(5)$ and $(6)$ both $\map {u'} x$ and $\map {v'} x$ maintain sign in the subintervals and vanish at one or both endpoints of each subinterval.
Let $J$ be one of the subintervals.
Let functions $\map {u'} x$, $\map {v'} x$ be zero at $\xi$, the right endpoint.
The quantity:
- $\map {y'} x - \dfrac {c_1 - a_1} {c - a}$
has to be either positive or negative.
Suppose it is positive in $J$.
From $(5)$, $u'$ is not negative.
Multiply both sides of $(10)$ by $u'$:
- $u'' u' \ge - 2 \AA \BB_1 u u'$
Integrating this from $x \in J$ to $\xi$, together with $\map {u'} \xi = 0$, yields:
- $-\map {u'^2} x \ge - 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}$
Then:
\(\ds \map {u'^2} x\) | \(\le\) | \(\ds 2 \AA \BB_1 \paren {\map {u^2} \xi - \map {u^2} x}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \AA \BB_1 \map {u^2} \xi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 \AA \BB_1 \exp \paren {4 \AA \paren {m + \map y x - \dfrac {a_1 \paren {c - x} + c_1 \paren {x - a} } {c - a} } }\) | Equation $(3)$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds 2 \AA \BB_1 e^{8 \AA m}\) | Lemma 2 | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {\map {u'^2} x} {\map {u^2} x}\) | \(\le\) | \(\ds 2 \AA \BB_1 e^{8 \AA m}\) | $u \ge 1$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4 \AA^2 \paren {\map {y'} x - \dfrac {c_1 - a_1} {c - a} }^2\) | \(\le\) | \(\ds 2 \AA \BB_1 e^{8 \AA m}\) | Equation $(5)$ |
Hence:
- $\forall x \in J: \size {\map {y'} x - \dfrac {c_1 - a_1} {c - a} } \le \sqrt {\dfrac {\BB_1} {2 \AA} } e^{4 \AA m}$
Similar arguments for aforementioned quantity being negative.
$\blacksquare$
Sources
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- 1912: S.N. Bernstein: Sur les équations du calcul des variations ("On the equations of the calculus of variations") (Ann. Sci. École Norm. Sup. Vol. 29: pp. 431 – 485)
- 1962: N.I. Akhiezer: The Calculus of Variations: $\S 1.9$: A Theorem of Bernstein
- 1963: I.M. Gelfand and S.V. Fomin: Calculus of Variations ... (previous) ... (next): $\S 1.4$: The Simplest Variational Problem. Euler's Equation
- 1978: A. Granas, R.B. Guenther and J.W. Lee: On a theorem of S. Bernstein (Pacific J. Math. Vol. 74, no. 1: pp. 67 – 82)